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SOVA2 [1]
3 years ago
10

Frankie and Maria paid to get into the carnival, and they had to pay a certain amount per game. Frankie played 5 games and spent

$15.50 altogether. Maria played 12 games and spent $26 altogether. How much does it cost to get into the carnival?

Mathematics
1 answer:
Natali [406]3 years ago
6 0

Answer:

It cost $8 per person to get into the carnival

Step-by-step explanation:

Let the entry cost be y

Let Cost of 1 game be x

Cost of 5 games = 5x

Cost of 12 games = 12x

Frankie played 5 games and spent $15.50 altogether.

So, 5x+y=15.50 --- A

Maria played 12 games and spent $26 altogether.

So, 12x+y=26 -- B

Plot the lines

5x+y=15.50 --- Black color

12x+y=26  --- Red color

Intersection point provides the solution

(x,y)=(1.5,8)

So, it cost $8 per person to get into the carnival

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Jocelyn and her children went into a bakery and she bought $9 worth of cookies and brownies. Each cookie costs $0.50 and each br
Yuliya22 [10]

Jace bought 4 cookies and 2 brownies from the bakery.

Let x represent the number of cookies and y represent the number of brownies.

Since he bought $9 worth of cookies and brownies. Each cookie costs $0.75 and each brownie costs $3. hence:

0.75x + 3y = 9      (1)

Also, he bought twice as many cookies as brownies, hence:

x = 2y

x - 2y = 0     (2)

Solving equation 1 and 2 simultaneously gives x = 4, y = 2

Hence Jace bought 4 cookies and 2 brownies from the bakery.

6 0
3 years ago
Helppppopleaseee????
rosijanka [135]

Answer:

LQ = 5

Median = 6

Step-by-step explanation:

\frac{n+1}{2} gives us 5.5, so we do the average of position 5 and 6. This gives us 6, so the median is 6.

Now we find the median of the bottom half of the numbers, which is 5, so the LQ is 5.

Now we find the median of the upper half of the numbers, which is 7, so the UQ is 7.

7 0
2 years ago
The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.9 minutes and a standard deviation of 2.9
Eva8 [605]

Answer:

a) 0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) 0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes

c) 0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 8.9 minutes and a standard deviation of 2.9 minutes.

This means that \mu = 8.9, \sigma = 2.9

Sample of 37:

This means that n = 37, s = \frac{2.9}{\sqrt{37}}

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

320/37 = 8.64865

Sample mean below 8.64865, which is the p-value of Z when X = 8.64865. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{8.64865 - 8.9}{\frac{2.9}{\sqrt{37}}}

Z = -0.53

Z = -0.53 has a p-value of 0.2981

0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

275/37 = 7.4324

Sample mean above 7.4324, which is 1 subtracted by the p-value of Z when X = 7.4324. So

Z = \frac{X - \mu}{s}

Z = \frac{7.4324 - 8.9}{\frac{2.9}{\sqrt{37}}}

Z = -3.08

Z = -3.08 has a p-value of 0.001

1 - 0.001 = 0.999

0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Sample mean between 7.4324 minutes and 8.64865 minutes, which is the p-value of Z when X = 8.64865 subtracted by the p-value of Z when X = 7.4324. So

0.2981 - 0.0010 = 0.2971

0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

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zhuklara [117]
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8 0
3 years ago
Solve: x/x+2 + 5/x-3 = 25/x2-x-6
Paraphin [41]

Answer:

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BUT x1=3 is not an answer because it doesnt respects the range on the original ecuation, so x2=-5 is the solution

Step-by-step explanation:

6 0
2 years ago
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